Suppose that $|A(n)|<B$ and $\lim_{s \to 0^{+}}\sum_{n=1}^{\infty}a_{n}n^{-s}=a$ where $A(x)=\sum_{n \leq x}a_{n}$. Then $$\lim_{x \to \infty}\sum_{n \leq x} a_{n}(1-\log{n}/\log{x})=a$$
What I did :
WLOG, we can take $a=0$. Through some calculations we get $$\alpha(s)=s\int_{1}^{\infty}\frac{A(n)}{n^{1+s}}dn$$ $$\sum_{n \leq x} a_{n}(1-\log{n}/\log{x})=\frac{1}{\log{x}}\int_{1}^{x}\frac{A(n)}{n}dn$$ where $\alpha(s)=\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$.
WTS : $\frac{1}{\log{x}}\int_{1}^{x}\frac{A(n)}{n}dn=o(1)$
By the Hardy-Littlewood theorem, we get $$\frac{1}{U}\int_{1}^{U}A(\log{n})dn=o(1)$$ Take $U=\log{x}$ and $n=\log{y}$. Then $$\frac{1}{\log{x}}\int_{1}^{x}\frac{A(\log{\log{y}})}{y}dy=o(1)$$ Hardy-Littlewood theorem.
Suppose that $I(s)=\int_{0}^{\infty}a(u)e^{-us}du=(a+o(1))s^{-\beta}$ and $a(u) \geq (a+o(1))(u+1)^{-\beta+1}$. Then $\int_{0}^{U}a(u)du=(a+o(1))U^{\beta}$
I need your help. Thanks in advance.
Hint:$$|\sum_{n \leq x} a_{n}(1-\log{n}/\log{x})|=|\frac{1}{\log{x}}\int_{1}^{x}\frac{A(t)}{t}dt|\le{1\over\log x}\int_{1}^{x}{Bdt \over t}=B$$
If signs of ${a_n}$ are alternating, then following would be an easier method.
$A(x)=o(1)$ by your assumption. So, $\alpha(s)$ is analytic for $\Re(s)\ge 0$ as $\lim\limits_{s\to 0+}\alpha(s)=0$. So it will suffice to show that, $\sum\limits_{n\le x}a_n\log n=o(\log x).$ But, applying $A(x)=o(1)$, $$\sum\limits_{n\le x}a_n\log n=A(x)\log x-\int_1^x{A(t)\over t}dt=o(\log x)$$